Mark is in a deep hole looking for treasure. he is standing 8 feet below the surface. he throws an old watch he found with an initial upward velocity of 23 ft/sec. how long until it lands outside the hole, having gone up and back down?

He trows a watch with an initial upward velocity: v o = 23 ft/s.
v = v o - g · t
At the highest point:
0 = 23 - 9.81 · t
9.81 t = 23
t = 23 : 9.81
t = 2.34 s
h = v o · t - g t² / 2
h = 23 · 2.34 - 9.81 · 2.34² / 2
h = 53.92 - 26.86 = 26.86 ft
26.86 ft - 8 ft = 18.86 ft
18.86 = g · t² / 2
t² = 18.86 · 2 : 9.81
t = 1.96 s
t ( total ) = 2.34 s + 1.96 s = 4.3 s
The watch will land outside the hole after 4.3 seconds.

21/2 or 5/2

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